Match List-I and List-II
For values of \( w \), \( x \), \( y \) and \( z \) in the nuclear fission reaction of uranium nucleus as:
\( {}_{92}\mathrm{U}^{w} + {}_{0}\mathrm{n}^{1} \rightarrow {}_{x}\mathrm{U}^{236} \rightarrow {}_{y}\mathrm{Ba}^{144} + {}_{36}\mathrm{Kr}^{z} + 3\,{}_{0}\mathrm{n}^{1} \)
| List-I | List-II | ||
|---|---|---|---|
| particulars | values | ||
| (A) | \( w \) | (I) | 56 |
| (B) | \( x \) | (II) | 235 |
| (C) | \( y \) | (III) | 89 |
| (D) | \( z \) | (IV) | 92 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
In every nuclear reaction the mass number (superscript here) and the atomic number (subscript) are conserved separately.
(A) \( w \) — this matches (II) 235. Mass number in the first step: \( w + 1 = 236 \), so \( w = 235 \).
(B) \( x \) — this matches (IV) 92. Atomic number in the first step: \( 92 + 0 = x \); the compound nucleus is still uranium.
(C) \( y \) — this matches (I) 56. Atomic number in the second step: \( 92 = y + 36 + 3(0) \), so \( y = 56 \) (barium).
(D) \( z \) — this matches (III) 89. Mass number in the second step: \( 236 = 144 + z + 3(1) \), so \( z = 89 \).
NCERT (2026-27):
Fission – “… An example of fission is when a uranium isotope \( {}^{235}_{92}\mathrm{U} \) bombarded with a neutron breaks into two intermediate mass nuclear fragments
\( \displaystyle {}^{1}_{0}\mathrm{n} + {}^{235}_{92}\mathrm{U} \rightarrow {}^{236}_{92}\mathrm{U} \rightarrow {}^{144}_{56}\mathrm{Ba} + {}^{89}_{36}\mathrm{Kr} + 3\,{}^{1}_{0}\mathrm{n} \qquad (13.10) \)”